Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 4, 2016 20:46:33 GMT
Vasco,
I'm puzzled by Figure 2: f = log z from your document.
We should be seeing the greatest rotation on the lever in the direction of $K$. Granted in this case we should probably see negative rotation, and Figure 2 (b) does show negative rotation in that direction, but it appears that the negative rotation of the levers increases with negative $\theta$ and reaches a maximum at or near $\pi/3 = -5pi/3$. Isn't this a contradiction to (31)?
I am seeing a similar pattern in the levers in my own diagram.
Gary
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Post by Admin on Apr 5, 2016 6:59:07 GMT
Vasco, I'm puzzled by Figure 2: f = log z from your document. We should be seeing the greatest rotation on the lever in the direction of $K$. Granted in this case we should probably see negative rotation, and Figure 2 (b) does show negative rotation in that direction, but it appears that the negative rotation of the levers increases with negative $\theta$ and reaches a maximum at or near $\pi/3 = -5pi/3$. Isn't this a contradiction to (31)? I am seeing a similar pattern in the levers in my own diagram. Gary Gary In (31) it says that "...$\widetilde{Q}$ rotates most rapidly and its size remains constant, when $Q$ moves in the direction of $\mathcal{K}$...". This means that as we travel on the LHS, from the white circle at 150 degrees to the opposite white circle at -30 degreees, along the diameter in the direction of $\mathcal{K}$, the lever on the image circles, (traversed in the direction north, centre, south) on the RHS, rotate anticlockwise through a larger angle than for any other diameter on the LHS. It seems to me that figure 2 is in agreement with this. It may not be correct overall but it does agree with (31). I have calculated the total amount of rotation for the six diameters and the maximum value is indeed along the north-south diameter. Similarly, in the direction of $-i\mathcal{K}$ on the LHS, the rotation on the RHS is zero and the amplification on the RHS is at its greatest, which again agrees with (31). A good way to think about (21) in my opinion, is as a shape at some point in the complex plane, with its image under $f$ in the image complex plane. We can now move the original point in any direction from its original position and watch what happens to its image. In one of these directions, $\mathcal{K}$, the image rotates most rapidly, without changing shape, and in the direction $-i\mathcal{K}$, at right angles to the direction of $\mathcal{K}$, the opposite is true, and there is no rotation but maximum magnification. In all the other directions both rotation and magnification change. As we move from the direction $\mathcal{K}$ towards the direction $-i\mathcal{K}$, the image rotation along the diameter decreases and the magnification along the diameter increases until we reach the direction $-i\mathcal{K}$, and if we keep going the reverse happens. We are interested here in movement along the spokes of the wheel, not around the circumference. Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 5, 2016 16:29:15 GMT
Vasco,
That all makes good sense. I had forgotten to take into account the rotation of the circles around the rim of the wheel due to rotation of $\xi$. A way to show just the rotation of the small shapes is to rotate the whole wheel by Pi/3 to bring the dashed arrow back into alignment with $K$. One could also draw a little arc of the right size and direction from the original angle of the bar to the new angle (taking Arg(f'(c)) into account). I will do this and post it, but first I have one or two other problems with my plot.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 6, 2016 7:36:37 GMT
Vasco, Here is my version of the Log plot, finally. I anticipate that the remainder of 17 will go more quickly. I see an advantage in your method of using the spreadsheet for calculations, as you have the ability to monitor the numbers. I had trouble plotting this until I obtained more succinct expressions for $\alpha$ and $\sigma$. nh.ch5.ex17.vi.log.pdf (117.86 KB) I see that the .pdf has a faint plot. I tried resaving and reloading it, but to no avail. I'll see what I can do. Meanwhile, I think the essentials can be read from it. Here, at least, is a .jpg file. Gary
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Post by Admin on Apr 6, 2016 15:37:54 GMT
Gary I see you've been burning the midnight oil! Your document and diagrams look good to me. Just one small thing. The dotted arrow at right angles to $\mathcal{K}$ should point in the direction $-i\mathcal{K}$, the direction in which the radius of the circles increases most rapidly, see (31) on page 240. In your diagrams it is pointing in the direction $+i\mathcal{K}$. I have modified my document regarding exercise 17 chapter 5 part (vi) again to add some explanation to the $f=z^m$ case and the Mobius case. Also I found a mistake in my diagrams for these cases and so I have amended them and re-published the document here. This still may not be the last revision. I am now trying to pull together the four or five documents I have produced over the last few weeks to produce a single document with my solution for all six parts of this exercise. Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 6, 2016 17:41:57 GMT
Gary I see you've been burning the midnight oil! Your document and diagrams look good to me. Just one small thing. The dotted arrow at right angles to $\mathcal{K}$ should point in the direction $-i\mathcal{K}$, the direction in which the radius of the circles increases most rapidly, see (31) on page 240. In your diagrams it is pointing in the direction $+i\mathcal{K}$. I have modified my document regarding exercise 17 chapter 5 part (vi) again to add some explanation to the $f=z^m$ case and the Mobius case. Also I found a mistake in my diagrams for these cases and so I have amended them and re-published the document here. This still may not be the last revision. I am now trying to pull together the four or five documents I have produced over the last few weeks to produce a single document with my solution for all six parts of this exercise. Vasco Midnight oil, yes, and for just one part of many of one problem in a text with many problems. No one can accuse Needham of making life easy for students. Thank you for the comments. But now that I have some idea of a valid approach, I'm looking forward to trying the z^m and Möbius problems. I look forward to your final document, but perhaps you should warn the readers that this is not an ordinary problem to be dispatched in a page or two! Thank you for the comments. I will amend the figure. Gary
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Post by Admin on Apr 6, 2016 18:37:02 GMT
Gary
Yes, I agree with you about this exercise. As you know when I first read it I wasn't sure what it was asking for, and then when I felt I did, it turned out to be an interesting, stimulating and difficult exercise - probably the most thought-provoking exercise in the book yet!
Also section IX in the book packs in so many challenging concepts in 7 pages, and with so many gaps for the reader to fill in! I feel that by persevering with the exercise I have come to really understand section IX and managed to fill in a lot of the gaps.
I wonder if the rest of the book is hiding more surprises like that one.
Vasco PS Right at the end of section IX I notice that Needham makes a reference to further developments for complex curvature in Chapter 12.
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